In an NPN transistor, why isn't a typical C-E voltage sufficient to overcome the internal resistance of NPN junction directly? Since we only need 0.7V to overcome the PN junction from the Base to Emitter, isn’t that the same as applying a similar (or slightly higher) voltage from Collector to Emitter to achieve the same outcome?
Thinking of it an electronic level, the key thing you’re trying to do is ‘push’ electrons into an area of negative charge – i.e. the Collector. I don’t see why the existing potential difference from Emitter to Collector doesn’t do this without a base potential being applied at all.
Best Answer
I suggest that you consider the physical structure of an NPN transistor. It is composed of a pair of PN junctions connected as
simulate this circuit – Schematic created using CircuitLab
When the base is disconnected, the upper junction is reverse-biased, so very little current flows.
Before you start objecting, you need to be aware that the two diodes are not perfectly isolated, and as soon as current starts flowing in the base-emitter junction, it affects the collector-base junction in such a way as to allow more current to flow. The details require a whole lot more space and math than is appropriate here.
Nonetheless, the open-base model is accurate at this level of discussion. For a more complete discussion see the Wiki page as a start
Finally, as a nitpick your opening assertion is simply wrong. Some current will flow from emitter to collector, although the amount is usually quite small. The name for this current is the "collector cutoff current", and is normally only of much importance for small-signal transistors.